Solve x²((8.9 − x)² − 64) = (71 × 8.9 − 79x)² for all real x.

Step-by-step solution

1. 1

   Clear the constants and expand both sides. First $71\cdot 8.9=631.9$. On the left,

   $(8.9-x)^2-64=x^2-17.8x+79.21-64=x^2-17.8x+15.21$

   $\text{LHS}=x^2(x^2-17.8x+15.21)=x^4-17.8x^3+15.21x^2$

   On the right, with $79^2=6241$, $2(631.9)(79)=99840.2$ and $631.9^2=399297.61$:

   $\text{RHS}=6241x^2-99840.2x+399297.61$

2. 2

   Move everything to one side. Subtracting the right side from the left:

   $x^4-17.8x^3+(15.21-6241)x^2+99840.2x-399297.61=0$

   $x^4-17.8x^3-6225.79x^2+99840.2x-399297.61=0$

   Call this $f(x)$. Note that squaring was already done for us here — the problem is the squared form — so every real root of this quartic is a genuine solution; there is nothing extraneous to discard.

3. 3

   Locate sign changes to bracket the real roots. Evaluating $f$ at a few points:

   $f(-90)\approx 1.877\times 10^{7},\quad f(-80)\approx 1.842\times 10^{6},\quad f(-70)\approx -7.779\times 10^{6}$

   $f(0)=-399297.61,\quad f(80)\approx -410738,\quad f(81)\approx 427422$

   There is one sign change between $-80$ and $-70$, and one between $80$ and $81$. Since a quartic has at most four real roots and the leading coefficient is positive, those two brackets hold one root each.

4. 4

   Refine both roots. Newton's method inside each bracket converges to

   $x_1\approx -78.5149849012,\qquad x_2\approx 80.4985052279$

   Substituting back into $x^2((8.9-x)^2-64)-(631.9-79x)^2$ gives residuals below $10^{-7}$ in magnitude, so both are genuine.

5. 5

   Prove there are no other real roots — this is the step that matters. On the whole interval $[0,80]$ the quartic never becomes positive: its largest value there is about $-3992$, attained near $x\approx 7.909$. Because $f<0$ across that entire stretch, no root can hide inside it. For $x<-78.51$ and $x>80.50$ the quartic is increasing in magnitude and stays positive, and on $(-78.51,0)$ it stays negative. So exactly two real roots exist; the remaining two are the complex pair $7.90824\pm 0.79772i$.

6. 6

   Beware the plausible-looking four-root answer. A commonly circulated solution lists $x\approx-86.19,\ 6.9982,\ 8.9387,\ 88.055$. Test the middle two directly: $f(6.9982)\approx -9205$ and $f(8.9387)\approx -10628$ — both far from zero, exactly as the $[0,80]$ bound predicts. Any time a quartic is reported with four real roots, plug at least one back in; the sign of $f$ on a whole interval is the cheapest disproof there is.